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Epsilon Delta Proof

  • Thread starter razored
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  • #1
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[SOLVED] Epsilon Delta Proof

Does this limit proof make total sense? Given : "Show that [tex]\lim_{x \rightarrow 2} x^{2} = 4[/tex]."

My attempt at it :[tex]0<|x^{2}-4|<\epsilon[/tex] which can also be written as [tex]0<|(x-2)(x+2)|<\epsilon[/tex].
[tex]0<|x-2|<\delta[/tex] where [tex]\delta > 0[/tex]. It appears that [tex]\delta = \frac {\epsilon}{x+2}[/tex] which is the conversion factor. Which then by substitution, [tex]0<|x-2|<\frac {\epsilon}{x+2}[/tex].

Here is the actual proof:
Choose [tex]\delta = \frac {\epsilon}{x+2}[/tex], given [tex]epsilon > 0[/tex] then if [tex]0<|x-2|<\frac {\epsilon}{x+2}[/tex] then [tex]0<|(x-2)(x+2)|<\delta[/tex].

Is this explanation coherent? I actually have a slight idea of what I wrote. Hopefully I am on the right path.
 
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Answers and Replies

  • #2
dynamicsolo
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One of the more rigorous mathematicians here may have more to say about this then I will, but I think you have the gist of it. The one safeguard that is imposed in epsilon-delta proofs when you have a non-constant upper limit on the inequality is to make that limit

min ( 1 , [tex]\frac {\epsilon}{x+2}[/tex] ) ,

since the ratio can blow up at x = -2 and we've placed no restriction on the permitted values of x...
 
  • #3
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One of the more rigorous mathematicians here may have more to say about this then I will, but I think you have the gist of it. The one safeguard that is imposed in epsilon-delta proofs when you have a non-constant upper limit on the inequality is to make that limit

min ( 1 , [tex]\frac {\epsilon}{x+2}[/tex] ) ,

since the ratio can blow up at x = -2 and we've placed no restriction on the permitted values of x...
I can't believe I forgot to mention that!
 
  • #4
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When you are doing [tex]\epsilon-\delta [/tex] proofs, you have to say define what epsilon and delta are i.e. [tex]\forall \epsilon >0, \exists \delta[/tex] such that etc.
 
  • #5
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Your [itex] \delta [/itex] should not depend on [itex] x [/itex]. As the post above mentions, the logic goes: for all [itex] \epsilon > 0 [/itex] there exists [itex] \delta > 0 [/itex] such that if [itex] |x-2| < \delta[/itex] then [itex] |x^2 - 4| < \epsilon [/itex]. Delta can, however, depend on epsilon. If |x-2| is bounded by [itex] \delta < 1 [/itex], then what is a bound on |x+2|?
 
  • #6
dynamicsolo
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Your [itex] \delta [/itex] should not depend on [itex] x [/itex].
Sorry -- quite so! It's been a while since I've looked at one of these proofs and I'm writing from someplace I don't have my books handy...
 
  • #7
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My second attempt : Given any [tex]\epsilon > 0[/tex], choose [tex]\delta = \frac {\epsilon} {|x+2|} (x\neq -2)[/tex] where [tex]\delta > 0[/tex], then [tex]|x-2||x+2|<\epsilon[/tex] whenever [tex]|x-2|<\delta = \frac {\epsilon} {|x+2|}[/tex].

Is this right?
 
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  • #8
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In my revision, I've been told that I did not prove anything. Is this a complete proof?
 
  • #9
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Let [tex]f(x)=x^2[/tex], [tex]x_{0}=2[/tex]. [tex]\forall \epsilon>0,\exists \delta>0[/tex] such that [tex]|x^2-4|<\epsilon[/tex], [tex]|x-2|<\delta[/tex]
Let [tex]\delta = \frac{\epsilon}{x+2}[/tex]
... continue from here. "..." [tex]<\epsilon[/tex]

OR let [tex]\delta=min\{ 1,\frac{\epsilon}{5} \}[/tex] and use the triangle inequality.
QED
 
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  • #10
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Substituting [tex]\delta = \frac {\epsilon}{x+2}[/tex] into [tex]|x-2|<\delta[/tex], we see that.. it works out? I do not know what to do from here.

Nor do I know the min way.
 
  • #11
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Continue from the proof,
Let [tex]\epsilon>0[/tex], let [tex]\delta = min\{\frac{\epsilon}{5}\} [/tex]

Assume that [tex]|x-2|<\delta[/tex]

By Triangle inequality,
[tex]|x^2-4|=|(x+2)(x-2)|\leq |x+2||x-2|[/tex]

Since [tex]|x-2|<\delta[/tex], then [tex]|x-2|<\epsilon/5[/tex]

Then for 0<x<3 such that [tex]|x+2|=5[/tex] **note: this is why you get the 5 as the denominator under epsilon.

Thus, [tex]|x^2-4|\leq |x+2||x-2|<5*\frac{\epsilon}{5}=\epsilon[/tex] QED
 

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